Coco dfs 或者 状压dp。...

C -- Coco

Time Limit：1s Memory Limit：64MByte

Submissions：148Solved：85

DESCRIPTION

Coco just learned a math operation call mod.Now,there is an integer $\mathrm{aa\; and\; nn\; integers\; b1,\dots ,bnb1,\dots ,bn.\; After\; selecting\; some\; numbers\; from\; b1,\dots ,bnb1,\dots ,bn\; in\; any\; order,\; say\; c1,\dots ,crc1,\dots ,cr,\; Coco\; want\; to\; make\; sure\; that\; amodc1modc2mod\dots modcr=0amodc1modc2mod\dots modcr=0\; (i.e.,\; aa\; will\; become\; the\; remainder\; divided\; by\; cici\; each\; time,\; and\; at\; the\; end,\; Coco\; want\; aa\; to\; become\; 00).\; Please\; determine\; the\; minimum\; value\; of\; rr.\; If\; the\; goal\; cannot\; be\; achieved,\; print\; -1-1\; instead.}$

INPUT

The first line contains one integer $\mathrm{T(T\le 5)T(T\le 5),\; which\; represents\; the\; number\; of\; testcases.\; For\; each\; testcase,\; there\; are\; two\; lines:\; 1.\; The\; first\; line\; contains\; two\; integers\; nn\; and\; aa\; (1\le n\le 20,1\le a\le 1061\le n\le 20,1\le a\le 106).\; 2.\; The\; second\; line\; contains\; nn\; integers\; b1,\dots ,bnb1,\dots ,bn\; (\forall 1\le i\le n,1\le bi\le 106\forall 1\le i\le n,1\le bi\le 106).}$

OUTPUT

Print $\mathrm{TT\; answers\; in\; TT\; lines.}$

SAMPLE INPUT

2 2 9 2 7 2 9 6 7

SAMPLE OUTPUT

2 -1

先对数组排序，从大到小，因为mod大的数，不会对MOD小的数有影响，但是因为余数比MOD小的数更大，所以因子数只会更多。机会只会更多，就像样例一样，先MOD 7，再MOD 2.

所以用dfs枚举每一个数，MOD和不MOD都选一次。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#define MY "H:/CodeBlocks/project/CompareTwoFile/DataMy.txt", "w", stdout
#define ANS "H:/CodeBlocks/project/CompareTwoFile/DataAns.txt", "w", stdout


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 1e6 + 20;
int a[maxn];
int n, ans;
int val;
void dfs(int selct, int re, int cur) {
    if (selct >= ans) return;
    if (re == 0) {
        ans = min(ans, selct);
        return;
    }
    if (cur > n) return;
    dfs(selct + 1, re % a[cur], cur + 1);
    dfs(selct, re, cur + 1);
}
void work() {
    scanf("%d%d", &n, &val);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n, greater<int>());
    ans = inf;
    dfs(0, val, 1);
    if (ans == inf) printf("-1
");
    else printf("%d
", ans);
}

int main() {
    freopen("data.txt","r",stdin);
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}

状压：dp[state]表示现在MOD了那位数字，然后枚举的时候，也是优先MOD了大的数字，再转移去小的数字。不允许先MOD小的再转移去MOD大的，就是2--3这条路径是不行的，因为2--3就是先MOD2，再MOD1得到，但是可以先MOD1，再MOD2得到。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
#define MY "H:/CodeBlocks/project/CompareTwoFile/DataMy.txt", "w", stdout
#define ANS "H:/CodeBlocks/project/CompareTwoFile/DataAns.txt", "w", stdout


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 2e6 + 20;
int dp[maxn];
int a[maxn];
int get(int val) {
    int ans = 0;
    while (val) {
        val &= (val - 1);
        ans++;
    }
    return ans;
}
int cnt[maxn];
void init() {
    int end = (1 << 20) - 1;
    for (int i = 0; i <= end; ++i) {
        cnt[i] = get(i);
    }
    return;
}
void work() {
    int ans = inf;
    int n, val;
    scanf("%d%d", &n, &val);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
    }
    sort(a, a + n, greater<int>());
    memset(dp, 0x3f, sizeof dp);
    for (int i = 0; i < n; ++i) {
        dp[1 << i] = val % a[i];
        if (dp[1 << i] == 0) ans = 1;
    }
    int end = (1 << n) - 1;
    for (int j = 1; j <= end; ++j) {
        if (cnt[j] >= ans) continue;
        for (int i = 0; i < n; ++i) {
            if (((1 << i) & j) != 0) continue;
            int state = j | (1 << i);
            if (cnt[state] >= ans) continue;
            if (dp[state] != inf) continue;
            dp[state] = dp[j] % a[i];
            if (dp[state] == 0) ans = cnt[state];
        }
    }
    if (ans == inf) printf("-1
");
    else printf("%d
", ans);
}

int main() {
    init();
    int t;
    scanf("%d", &t);
    while (t--) work();
    return 0;
}